A commercial rubbing alcohol contains 37. g of isopropanol and 0.207 g of sucrose octaacetate in each 51. -mL portion. Calculate the weight/volume percent concentration for each component. Be sure each of your answer entries has the correct number of significant figures.

Question

Part 1 of 2

%

weight/
volume

isopropanol

Part 2 of 2

%

weight/
volume

sucrose octaacetate

Answer 1

Part 1:

First, calculate the weight of isopropanol in the 51 mL portion:
37 g isopropanol / 51 mL = 0.725 g isopropanol/mL

Next, convert the weight of isopropanol to a weight/volume percent concentration:
(0.725 g isopropanol/mL) * 100 = 72.5%

Therefore, the weight/volume percent concentration of isopropanol in the rubbing alcohol is 72.5%.

Part 2:

First, calculate the weight of sucrose octaacetate in the 51 mL portion:
0.207 g sucrose octaacetate / 51 mL = 0.00406 g sucrose octaacetate/mL

Next, convert the weight of sucrose octaacetate to a weight/volume percent concentration:
(0.00406 g sucrose octaacetate/mL) * 100 = 0.806%

Therefore, the weight/volume percent concentration of sucrose octaacetate in the rubbing alcohol is 0.806%.